$x^5+x-1=0\;,\;x\in\mathbb{R}$

Question

Given the following equation $$x^5+x-1=0\;,\;x\in\mathbb{R}$$

How to prove that (unevaluated)

$$x=\dfrac13\left(-1+\sqrt[3]{\dfrac{25}2-\dfrac{3\sqrt{69}}2}+\sqrt[3]{\dfrac12\left(25+3\sqrt{69}\right)}\right).$$

Any hint would be appreciated.

$\begingroup$ Hint: you can factor the polynomial. $\endgroup$
– Nishant
Jun 29, 2014 at 18:47 — Nishant, Jun 29, 2014 at 18:47

Mathmo123 · Accepted Answer · 2014-06-29 18:52:32Z

9

Hint: $x^5 + x - 1 = (x^3 + x^2 -1)(x^2-x+1)$. Solving the first of these will give your solution.

answered Jun 29, 2014 at 18:52

Mathmo123

23k3 gold badges47 silver badges83 bronze badges

4

$\begingroup$ And it should be established that the other four roots are not real, so that we may conclude $x$ is this number with certainty. $\endgroup$
– 2'5 9'2
Jun 29, 2014 at 19:10

Add a comment |

1 Answer 1